eigenvectors-eigenvalues.wiki (2749B)
1 == Eigenvectors & eigenvalues == 2 let A be n × n, $x \in \Re^n$ is an eigenvector of A if x ≠ 0 and $\exists \lambda \in \Re$ such that $Ax = \lambda x$ 3 4 x is eigenvector with corresponding eigenvalue λ. 5 6 Is a given vector $u \in \Re^n$ an eigenvector of a given A (n × n)? 7 * Do $Au$, check if result is a multiple of u. 8 9 Is a given λ an eigenvalue of A? 10 * $\exists x \ne 0$ such that $Ax - \lambda x = 0 \leftrightarrow (A-\lambda I_n)x = 0$ with nontrivial solutions. 11 12 The solution set of $(A-\lambda I_n)x = 0$ is the eigenspace corresponding to λ. 13 14 How to find a basis for the eigenspace of a given λ? 15 1. calculate matrix for $A-\lambda I_n$ where n is the number of rows or columns of A 16 2. reduce matrix to reduced echelon form 17 3. express solutions in parametric form (basic variables in terms of free variables) 18 4. basis for eigenspace is the set of the coefficients 19 20 If λ = 0, then Ax = 0 has a nontrivial solution (and A is _not_ invertible). 21 22 Eigenvectors corresponding to distinct eigenvalues are linearly independent. 23 24 === Determinant === 25 Geometric interpretation: let $A = [a_1 \; a_2]$. then the determinant (absolute value) is the surface area (or volume in 3D): 26 27 {{file:img/determinant-geometric-diagram.png|Determinant geometric diagram}} 28 29 Let A (n × n). A ~ U without scaling and using _r_ row interchanges. then $\det A = (-1)^r u_{11} \times \dots \times u_{nn}$ 30 31 A is invertible iff $\det A \ne 0$ 32 33 $\det AB = (\det A)(\det B)$ 34 35 λ is an eigenvalue of A iff $\det (A-\lambda I) = 0$ (the characteristic equation of A) 36 37 The eigenvalues of A (n × n) are the solutions for λ. Multiplicity is the number of solutions for λ. 38 39 === Similarity === 40 given A and B (n × n), A is similar to B if ∃p s.t. $A = PBP^{-1}$ 41 42 If A and B are similar, then they have the same characteristic polynomials (and the same eigenvalues with the same multiplicities) 43 44 === Diagonalization === 45 A is diagonalizable if A is similar to a diagonal matrix. 46 47 Diagonalization Theorem: A (n × n) is diagonalizable iff A has n linearly independent eigenvectors (the eigenbasis for $\Re^n$) 48 49 $A = P D P^{-1} \leftrightarrow$ columns of P are linearly independent eigenvectors, and the diagonal values of D are the eigenvalues corresponding to the eigenvectors in P. 50 51 How to diagonalize a matrix: 52 1. Find eigenvalues of A 53 2. Find n = λ linearly independent eigenvectors 54 3. Construct $P = \begin{bmatrix} p_1 & p_2 & \ldots & p_n \end{bmatrix}$ 55 4. Construct D from the corresponding eigenvalues on the diagonal. Order of eigenvalues must match the order for columns of P. 56 5. Check $A = p D p^{-1} \leftrightarrow Ap = pD$ (if p is invertible) 57 58 If A (n × n) has n distinct eigenvalues, it is diagonalizable. 59